# A projectile is shot at an angle of (5pi)/12  and a velocity of  2 m/s. How far away will the projectile land?

Mar 7, 2016

$\approx 0.2 m$

#### Explanation:

vertcal component of the velocity of projection $u \sin \alpha = 2 \cdot \sin \left(5 \frac{\pi}{12}\right) = 2 \cdot 0.97 = 1.94 m {s}^{-} 1$
if t be the time of flight
then
$0 = u \sin \alpha t - \frac{1}{2} g \cdot {t}^{2}$
$t = 2 u \sin \frac{\alpha}{g} = 2 \cdot \frac{1.94}{10} = 0.388 s$
The horizontal displacement of the projectile $u \cos \alpha t = 2 \cdot \cos \left(5 \frac{\pi}{12}\right) \cdot 0.388 \approx 0.2 m$